To find the position of the center of mass of the system, we need to first calculate the coordinates of the center of mass along both the x-axis and y-axis.
Let's begin by finding the coordinates of the center of mass along the x-axis:
We can use the formula:
x_cm = (m1x1 + m2x2 + m3x3 + m4x4) / (m1 + m2 + m3 + m4)
where m1, m2, m3, and m4 are the masses of the point masses at each corner, and x1, x2, x3, and x4 are the x-coordinates of each point mass.
In this case, we have:
m1 = 2kg, x1 = 0cm
m2 = 4kg, x2 = 2cm
m3 = 6kg, x3 = 2cm
m4 = 4kg, x4 = 0cm
Substituting these values into the formula, we get:
x_cm = (2kg x 0cm + 4kg x 2cm + 6kg x 2cm + 4kg x 0cm) / (2kg + 4kg + 6kg + 4kg)
x_cm = 24 / 16
x_cm = 1.5cm
Therefore, the x-coordinate of the center of mass is 1.5cm.
Now let's find the coordinates of the center of mass along the y-axis:
We can use the formula:
y_cm = (m1y1 + m2y2 + m3y3 + m4y4) / (m1 + m2 + m3 + m4)
where m1, m2, m3, and m4 are the masses of the point masses at each corner, and y1, y2, y3, and y4 are the y-coordinates of each point mass.
In this case, we have:
m1 = 2kg, y1 = 0cm
m2 = 4kg, y2 = 0cm
m3 = 6kg, y3 = 2cm
m4 = 4kg, y4 = 2cm
Substituting these values into the formula, we get:
y_cm = (2kg x 0cm + 4kg x 0cm + 6kg x 2cm + 4kg x 2cm) / (2kg + 4kg + 6kg + 4kg)
y_cm = 20 / 16
y_cm = 1.25cm
Therefore, the y-coordinate of the center of mass is 1.25cm.
So the center of mass of the system is located at the point (1.5cm, 1.25cm) from the corner.
Diagram A shows a negatively charged conducting rod placed near a light polystyrene ball that is suspended from the ceiling by an insulating thread .Diagram B shows what happens when the ball touches the rod. (a) Explain why the ball is displaced vertically in Diagram A (b)Explain what happens after the ball has been allowed to touch the rod (c)Give a reason why the ball has to be coated with a conducting material such as graphite (d) Explain why the polystyrene ball is suspended by an insulated thread and not by a conducting wire
When a negatively charged conductive rod is brought close to a lightweight Styrofoam ball suspended from the ceiling by insulating threads (see Figure A), the electrons in the ball are repelled by the rod's negative charge.
What happens when the ball hits the pole?As a result, the electrons in the sphere move away from the rod and spread unevenly over the surface of the sphere. This makes the side of the sphere closest to the stick positively charged and the opposite side negatively charged.When the ball touches the negatively charged wand (see Figure B), the negatively charged electrons in the wand repel the negatively charged electrons in the ball and move away from the contact point. This creates a charge imbalance on the surface of the ball, with excess positive charge on one side and negative charge on the other.As a result of this charge imbalance, the ball experiences an electrostatic force and moves vertically away from the rod. The direction of displacement depends on the relative magnitudes of the electrostatic force and the weight of the sphere. If the electrostatic force is stronger than the weight of the ball, the ball will move up. If the weight of the ball is stronger than the electrostatic force, the ball will move down. In both cases the ball moves vertically.For more information on electrostatic force kindly visit to
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Please answer this question, I’ll give brainliest if it’s correct!
Q5. If you heat a 0.45 kg sample of an unknown substance to a temperature of 80 °C then quickly transfer it into a beaker that contains 0.70 kg of water at 15 °C (the specific heat capacity of the water is 4200 J/kg•C), the highest temperature that the sample and water will attain is 21 °C. What is the heat capacity of the sample? [4 marks]
In which of the following substances will sound travel the fastest?
air
O iron
water
Answer:
iron
Explanation:
iron is a solid where the particles are closer together, so sound will vibrate the particles faster than in a liquid or gas where the particles are further spaced out.
Answer:
iron
Explanation:
iron is a solid where the particles are closer together, so sound will vibrate the particles faster than in a liquid or gas where the particles are further spaced out.
One elevator arrangement includes the passenger car, a counterweight, and two large pulleys, as shown in Figure 11-50. Each pulley has a radius of 1.2 m and a moment of inertia of 380 kg•m². The top pulley is driven by a motor. The elevator car plus passengers has a mass of 3100 kg, and the counterweight has a mass of 2700 kg. If the motor is to accelerate the elevator car upward at 1.8 m/s², how much torque must it generate?
The torque generated by the motor, accelerating with 1.8 m / Of the elevator is, 18372 Nm.
What is torque?
Torque is the term for the force that rotates an object about any axis. It is equivalent to the product of force and perpendicular distance in mathematical terms.
Given: Radius of pulley (r) = 1.2 meters;
Mass of counterweight (W) = 2700 kg;
Mass of elevator plus passenger (M) = 3100 kg;
Moment of inertia of pulley (I) = 380 kg
To calculate torque:
Now if the rope tension is then
T₁ = 3100 × (9.8 + 1.8)
T₁ = 35960 N
Now Angular acceleration (α) = a / r
α = 1.8 / 1.2
α = 1.5 rad / sec
If T₂ is the counterweight tension then
T₂ = 2700 × (9.8 - 1.8) N
T₂ = 21600 N
If T is the total torque then
T = 2 × 380 × 1.5 + (35960 - 21600) ×1.2
T = 18372 Nm
Therefore, the total torque that is generated through the given measurement is, 18372 Nm.
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lucy finished 1/4 of her homework at an average speed of 15 questions per hour then she finished the remaining 45 questions at another speed. if the total time spent on the homework was 2.5 hours, what was the amount of time she spent on the remaining 45 questions
Therefore, Lucy spent 1.875 hours on the remaining 45 questions.
How is the amount of time she spent on the remaining 45 questions?Let's start by finding the total number of questions in Lucy's homework. If she finished 1/4 of her homework at an average speed of 15 questions per hour, then the total number of questions must be:
1/4 x Total number of questions = Number of questions finished at 15 questions per hour
1/4 x Total number of questions = 15 questions per hour
Solving for the total number of questions, we get:
Total number of questions = 60 questions
Now we know that Lucy finished 60 - 45 = 15 questions at an average speed of x questions per hour. Let's use the formula:
time = distance / speed
to find the amount of time she spent on the remaining 45 questions.
For the first part of the homework, Lucy spent:
time = distance / speed
time = 15 questions / hour
time = 1/4 x 2.5 hours
time = 0.625 hours
So, she spent 0.625 hours on the first 15 questions.
For the remaining 45 questions, we have:
time = distance / speed
time = 45 questions / x questions per hour
We know that the total time spent on the homework was 2.5 hours, so:
0.625 hours + 45 questions / x questions per hour = 2.5 hours
Solving for x, we get:
x = 18 questions per hour
Now we can use this speed to find the time spent on the remaining 45 questions:
time = distance / speed
time = 45 questions / 18 questions per hour
time = 2.5 hours - 0.625 hours
time = 1.875 hours
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A particle was moving in a straight line with a constant acceleration. If the particle
covered 17 m in the 2nd second and 46 m in the 9th and 10th seconds, calculate its
acceleration a and its initial velocity vo.
Answer:
Acceleration (a) = 3 m/s^2
Initial velocity (vo)= 7 m/s.
Explanation:
We can use the equations of motion to solve this problem. Let's start with the second equation of motion:
d = vot + (1/2)at^2
where d is the displacement, vo is the initial velocity, a is the acceleration, and t is the time.
Using this equation for the 2nd second, we have:
17 = vo(2) + (1/2)a(2^2)
17 = 2vo + 2a
Using the same equation for the 9th and 10th seconds, we have:
46 = vo(10) + (1/2)a(10^2) - vo(9) - (1/2)a(9^2)
46 = 10vo + 50a - 9vo - 40.5a
46 = vo + 9.5a
Now we have two equations with two unknowns (vo and a). We can solve for one variable in terms of the other and substitute into the other equation. For example, we can solve the first equation for vo:
2vo = 17 - 2a
vo = (17 - 2a)/2
Now we can substitute this expression for vo into the second equation:
46 = [(17 - 2a)/2] + 9.5a
Solving for a, we get:
a = 3 m/s^2
Now we can use the expression for vo to find its value:
vo = (17 - 2a)/2
vo = (17 - 2(3))/2
vo = 7 m/s
Therefore, the acceleration of the particle is 3 m/s^2 and its initial velocity is 7 m/s.
Answer:
Acceleration (a) = 3 m/s^2
Initial velocity (vo)= 7 m/s.
Explanation:
We can use the equations of motion to solve this problem. Let's start with the second equation of motion:
d = vot + (1/2)at^2
where d is the displacement, vo is the initial velocity, a is the acceleration, and t is the time.
Using this equation for the 2nd second, we have:
17 = vo(2) + (1/2)a(2^2)
17 = 2vo + 2a
Using the same equation for the 9th and 10th seconds, we have:
46 = vo(10) + (1/2)a(10^2) - vo(9) - (1/2)a(9^2)
46 = 10vo + 50a - 9vo - 40.5a
46 = vo + 9.5a
Now we have two equations with two unknowns (vo and a). We can solve for one variable in terms of the other and substitute into the other equation. For example, we can solve the first equation for vo:
2vo = 17 - 2a
vo = (17 - 2a)/2
Now we can substitute this expression for vo into the second equation:
46 = [(17 - 2a)/2] + 9.5a
Solving for a, we get:
a = 3 m/s^2
Now we can use the expression for vo to find its value:
vo = (17 - 2a)/2
vo = (17 - 2(3))/2
vo = 7 m/s
Therefore, the acceleration of the particle is 3 m/s^2 and its initial velocity is 7 m/s.
Explanation:
(a) Calculate the force (in N) the woman in the figure below exerts to do a push-up at constant speed, taking all data to be known to three digits. (You may need to use torque methods from a later chapter.) 401.15
(b)How much work (in J) does she do if her center of mass rises 0.260 m?
(c) What is her useful power output (in W) if she does 30 push-ups in 1 min? (Should work done lowering her body be included? See the discussion of useful work in Work, Energy, and Power in Humans.)
the force the woman exerts to do a push-up at constant speed is 333 N.
the work the woman does is 152 J.
her useful power output is 76 W.
(a) To calculate the force the woman exerts to do a push-up, we need to use torque methods. The woman is doing a push-up at constant speed, which means that the net torque on her body is zero. The only torque acting on her body is due to her weight W, which acts at the center of mass of her body. The distance between her center of mass and her hands is 0.76 m, and the angle between her body and the horizontal is 45 degrees.
The torque due to her weight about her hands is given by:
τ = r x W = (0.76 m) x (cos 45°)(W)
where r is the distance between her hands and her center of mass and cos 45° is the component of the distance perpendicular to the weight vector. Since the woman is at constant speed, the torque she exerts about her hands must be equal and opposite to the torque due to her weight. Therefore:
τ = (0.76 m)(cos 45°)(W) = (1/2)(W)(0.76 m)
Solving for W, we get:
W = 2(τ/0.76 m) = 2[(0.5)(mg)(0.76 m)/(0.76 m cos 45°)] = 333 N
Therefore, the force the woman exerts to do a push-up at constant speed is 333 N.
(b) The work the woman does is equal to the change in her potential energy as her center of mass rises. The woman's mass is not given, so we will assume a value of 60 kg. The gravitational potential energy of the woman is given by:
U = mgh
where m is the mass of the woman, g is the acceleration due to gravity (9.81 m/s^2), and h is the height her center of mass rises (0.26 m). Therefore:
U = (60 kg)(9.81 m/s^2)(0.26 m) = 152 J
Therefore, the work the woman does is 152 J.
(c) The useful power output of the woman is the work she does per unit time, taking into account the work done in lowering her body. Each push-up involves two phases: lifting her body and lowering her body. When she lowers her body, the work done is negative, as the force she exerts is in the opposite direction to the displacement. The work done in lowering her body is equal to the work done in lifting her body, so the total work done in one push-up is zero.
The woman does 30 push-ups in 1 minute, which means she does one push-up every 2 seconds. Therefore, the useful power output of the woman is:
P = (152 J)/(2 s) = 76 W
Therefore, her useful power output is 76 W.
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2. There are many factors that play a role in body image. How might a person's culture
and family background shape their body image? You may choose to examine your own
ethnicity or discuss a variety of different ethnicities
A person's culture and family background can significantly shape their body image. Different cultures have unique standards of body ideals, which can influence how people perceive their own bodies and others.
What is culture?
In some cultures, a larger body size is considered desirable and associated with prosperity, fertility, and beauty, while in other cultures, a thinner body size is considered more attractive and associated with success, health, and discipline.
Family background can also play a role in shaping body image. For instance, if family members constantly make negative comments about weight or appearance, this can negatively impact a person's body image and self-esteem. Additionally, if a person grows up in a family where unhealthy eating habits are normalized or encouraged, this can contribute to the development of unhealthy body image and eating behaviors.
Overall, it is important to recognize the influence that culture and family background can have on body image, and to promote a more diverse and inclusive standard of beauty that celebrates different body types and sizes.
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12) The notion of task-appropriate processing implies that if you are preparing for a quiz you
should...
As part of a movie stunt, a full-size remote-controlled car is driven horizontally off a 9.00 m tall cliff at 24.40 m/s. How far (Δx) from the bottom of the cliff does the car land?
Explanation:
Find the time it takes to hit the bottom....then multiply this time by the horizontal velocity .......
Time to hit bottom :
d = 1/2 at^2
9 m = 1/2 (9.81 m/s^2) (t^2) shows t = 1.35 s
Now the horizontal displacement is
x = rate * time = 24.40 m/s * 1.35 s = 33.1 m
If I want to convert 6,785 mg into grams. I would need to move the decimal places to the
You divide by 1000 so it would be 6.785g
A 2.99 x 10-6 C charge is moving
perpendicular (90°) to the Earth's
magnetic field (5.00 x 10-5 T). If the
force on it is 2.14 x 10-8 N, how fast is
it moving?
[?] m/s
Velocity (m/s)
Enter
The charged particle is moving at a speed of 1.424 x 10³ m/s.
What is magnetic field?In the vast expanse of space, a magnetic force can be detected within a region known as a magnetic field. This field is generated by either a magnet, an electric charge in motion, or an electric field undergoing change. To visually represent this force, it is depicted through directional field lines which illustrate the direction of the force on an imaginary magnetic pole situated at any point in space.
The intensity of this magnetic field is quantified by its measurement in units called Tesla (T).
Equation:The force on a charged particle moving perpendicular to a magnetic field is given by the equation:
F = Bqv
where F is the force on the charge, B is the magnetic field strength, q is the charge of the particle, and v is its velocity.
Rearranging this equation, we can solve for v:
v = F / (B*q)
Substituting the given values, we get:
v = (2.14 x 10⁻⁸ N) / (5.00 x 10⁻⁵T * 2.99 x 10⁻⁶ C)
v = 1.424 x 10³ m/s
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Answer:142
Explanation: on acellus, and maybe on what ever else you are on, but i do know it's right on acellus!! :)
If the universe were to suddenly begin shrinking rather than continue expanding, how would it affect the cosmic microwave background radiation?
A. It would decrease in temperature.
B. It would blue-shift.
C. It would red-shift.
D. It would increase in temperature.
C. It would red-shift. The cosmic microwave background radiation (CMB) is the leftover radiation from the Big Bang.
What is radiation?Radiation is the process by which energy is emitted from a source and travels through a medium or space. It is a form of energy that can be found in a variety of forms, such as light, heat, and sound. Radiation can also be used to refer to the release of particles or electromagnetic waves from an atomic nucleus during radioactive decay. These particles and waves can be harmful to humans and animals if they are too strong or long lasting. Radiation is also used in medical, industrial, and scientific applications to diagnose and treat diseases, sterilize products, and to provide energy for power plants.
It is composed of a low-energy form of light called microwaves. If the universe were to start shrinking, this microwaves would be stretched out (red-shifted) as the universe gets smaller. This red-shifting would cause the CMB to become less energetic, and therefore decrease in temperature.
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The cylinder of a heat engine is filled with an air-fuel mixture. Which property of gases is essential to heat engines ability to do work?
Answer: Pressure of gases.
Explanation:
In a heat engine, the air-fuel mixture is ignited, which causes an increase in pressure of the gases inside the cylinder. This pressure pushes the piston, which is connected to a crankshaft, causing it to rotate and do work. Therefore, pressure is an essential property of gases for the ability of a heat engine to do work.
. If the same quantity of is supplied to P and Q and P has a temperature rise twice that of Q and has a mass that is half of the mass of Q. Find the ratio of the specific heat capacity of P to Q
The ratio of the specific heat capacity of P to Q is 1:2, or cP/cQ = 1/2.
Specific heat is the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). It is a physical property that helps characterize the thermal behavior of materials.
Let's use the formula for heat energy:
Q = mcΔT
where Q is the heat energy supplied, m is the mass of the object, c is the specific heat capacity, and ΔT is the change in temperature.
We can write two equations for P and Q using this formula:
For P: Q = (1/2)mP * cP * 2ΔT
For Q: Q = mQ * cQ * ΔT
We know that both P and Q receive the same amount of heat energy, so we can equate the two equations:
(1/2)mP * cP * 2ΔT = mQ * cQ * ΔT
Simplifying this expression, we get:
cP/cQ = mQ/(2mP)
We are given that P has half the mass of Q, so mP = (1/2)mQ. Substituting this into the expression above, we get:
cP/cQ = mQ/(2(1/2)mQ) = 1/2
Therefore, the ratio of the specific heat capacity of P to Q is 1:2, or cP/cQ = 1/2.
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A particular star is d = 56.1 light-years (ly) away, with a power output of P = 3.00 x 1026 W. Note that one light-year is the distance traveled by the light through a vacuum in one year. (a) Calculate the intensity of the emitted light at distance d (in nW/m2). nW/m2 (b) What is the power of the emitted light intercepted by the Earth (in kW)? (The radius of Earth is 6.37 x 10 m.) kw What If?
(a) Intensity of emitted light at distance d is approximately 1.95 nW/m². (b) Power of emitted light intercepted by Earth is approximately 3.67 kW.
How did you calculate the intensity of emitted light in part (a)?The intensity of emitted light at distance d was calculated using the inverse square law, which states that intensity is inversely proportional to the square of the distance. I used the equation I = P/(4πd²) to calculate the intensity, where P is the power output of the star and d is the distance to the star.
How did you calculate the power of emitted light intercepted by Earth in part (b)?To calculate the power of emitted light intercepted by Earth, I first calculated the surface area of a sphere with a radius equal to the distance between the Earth and the star. Then, I multiplied the intensity of the emitted light at that distance by the surface area of the sphere to get the total power of emitted light intercepted by Earth.
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v=√gr tan 31.0 grados
The banking angle is the angle that the surface makes with the horizontal, or the angle of inclination.
v = √rg tanθ
tanθ=v²/rg
The relation gives the angle of banking of the cyclist going round the curve. Here v is the speed of the cyclist, r is the radius of the curve, and g is the acceleration due to gravity.
The banking angle is the angle that the surface makes with the horizontal, or the angle of inclination. The normal force acting on the car while travelling through such a curving road has a horizontal component. The centripetal force needed to prevent skidding is provided by this component.
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Use the heating curve (Figure 1) to answer the question.
Temperature (C)
200
150
100
50
0
-50
-100
-150
0 200 400 600 800 1000 1200
Energy (J)
What do the horizontal sections represent?
They represent where no energy was added.
They represent where phase changes occur.
MacBook Air
They represent where particles moved faster.
They represent where the temperature changed.
They represent where phase changes occur.
what type of of electrical energy is produced by batteries
The type of electrical energy produced by batteries is chemical potential energy. One or more cells that transform chemical energy into electrical energy make up a battery.
What is electrical energy?A type of energy known as electrical energy is connected to the movement of electric charge. It is the energy that is transported across a conductor by moving electrons in an electric circuit.
Direct current (DC) batteries generate electrical energy. In contrast to alternating current (AC), which occasionally flips direction, direct current is a type of electrical energy that only flows in one direction through a circuit. A chemical reaction that takes place inside the battery itself produces the electrical energy that a battery generates.
An electrical potential difference between the battery's positive and negative terminals is produced by a chemical reaction that occurs inside each cell. When a circuit is attached to a battery, electrons move from the battery's negative terminal through the circuit and out to the positive terminal, causing an electrical current to flow.
Little electronic devices like calculators and lamps can be powered by batteries, as well as bigger applications like electric vehicles and backup power systems. Batteries have a finite capacity, though, and over time as their chemical reactions wear out, they will eventually get depleted.
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In which of the following scenarios is the left hemisphere of the brain primarily needed?
The left hemisphere of the brain is primarily needed in scenario, Solving a complex mathematical problem. Option a is correct.
The brain is divided into two hemispheres, left and right, and they are specialized for different cognitive functions. The left hemisphere of the brain is primarily responsible for language processing, logical reasoning, and analytical thinking. Solving a complex mathematical problem involves logical reasoning, analytical thinking, and the use of language, all of which are primarily controlled by the left hemisphere of the brain.
Mathematical problems often require precise calculations, sequencing of steps, and the use of symbols and formulas, all of which require a strong left-brain function. In contrast, appreciating a work of art, listening to music, and recognizing facial expressions are all more complex perceptual and emotional processes that involve the right hemisphere of the brain. Option a is correct.
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--The complete question is, In which of the following scenarios is the left hemisphere of the brain primarily needed?
a. Solving a complex mathematical problem.
b. Appreciating a work of art
c. Listening to music
d. Recognizing facial expressions.--
Suppose studentsA,B,C, and D are running a 100meter race student A takes12.5 seconds ,student B takes 10 seconds , students C takes 14 seconds in student D takes 15 seconds . calculate their respective velocities
Answer:
Va=8m/s Vb =10m/s Vc=7.14m/s Vs=6.67
Explanation:
#student A runs 100m in 12.5 seconds so in average he run 8 m in every second
#student B runs 100m in 10 seconds those means I'm average the student run 10m in each second.
#student C runs 100m in 14seconds so in average he/she run approximately 7.14m in every second.
#student D run 100m in 15secends so in average the student run approximately 6.67m in each second.
4. Two identical test tubes are filled with equal volumes of water and
mercury. Which of the following statements is true?
The weight of each liquid is the same.
The bottom area of each test tube is the same.
The pressure at the bottom of each test tube is the same.
All the above.
A bag of sand weighing 5kg is suspended from the lower end of a rope. The bag is initially at rest. A 20.0 g bullet is fired at the bag with horizontal velocity of 650 ms¹, strikes the block, and exits with 100 ms. To what vertical height will the block be raised?
Please help me in this exercise.
A. We can see here that the girl has kinetic energy with respect to the escalator.
B. The kinetic energy doesn't depend on the chosen reference.
What is kinetic energy?Kinetic energy is the energy that an object possesses due to its motion. Any object that is in motion has kinetic energy, which is determined by its mass and velocity. The formula for kinetic energy is KE=1/2mv^2, where KE is the kinetic energy, m is the mass of the object, and v is the velocity or speed of the object.
Kinetic energy is a scalar quantity, meaning it has only magnitude and no direction. The unit of kinetic energy is Joules (J) in the International System of Units (SI).
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In this problem you will go through a simplified version of Rutherford’s calculation of the size of the gold nucleus. Suppose a piece of gold foil that is 0.010 cm thick and whose area is 1 cm x 1 cm is used in the experiment. It is observed that 99.93% of all -particles go through undeflected. The density of gold is 19,300 kg/m3. Determine the radius of the gold nucleus. Hint: first calculate the total number of gold atoms in the foil
A simplified version of Rutherford’s calculation of the size of the gold nucleus. A piece of gold foil that is 0.010 cm thick and whose area is 1 cm x 1 cm is used in the experiment. It is observed that 99.93% of all particles go through undeflected. The density of gold is 19,300 kg/m3.
Rutherford's experiment involved firing alpha particles (helium nuclei) at a thin sheet of gold foil to study the structure of atoms. Based on the results of this experiment, Rutherford was able to deduce that atoms have a small, dense nucleus at their center.
In this problem, we will go through a simplified version of Rutherford's calculation of the size of the gold nucleus.
First, we need to calculate the total number of gold atoms in the foil. We know that the foil is 0.010 cm thick and has an area of 1 cm x 1 cm, so its volume is
V = thickness x area = 0.010 cm x (1 cm x 1 cm) = 0.010 [tex]cm^{3}[/tex]
The density of gold is 19,300 kg/[tex]m^{3}[/tex], which is equivalent to 19.3 g/[tex]cm^{3}[/tex]Therefore, the mass of the gold foil is
m = density x volume = 19.3 g/[tex]cm^{3}[/tex] x 0.010 [tex]cm^{3}[/tex] = 0.193 g.
The molar mass of gold is 197 g/mol, so the number of gold atoms in the foil is
N = (0.193 g) / (197 g/mol) x (6.022 x [tex]10^{23}[/tex] atoms/mol) = 1.86 x [tex]10^{21}[/tex] atoms
Next, we need to determine the fraction of alpha particles that are deflected by the gold nucleus. We are told that 99.93% of all alpha particles go through undeflected, which means that only 0.07% of the alpha particles are deflected. This is a very small fraction, which suggests that the size of the gold nucleus must be very small compared to the size of the atom.
Assuming that the alpha particles are deflected only by the gold nucleus and not by the electrons, we can use the principle of conservation of momentum to estimate the size of the gold nucleus. When an alpha particle approaches the gold nucleus, it experiences a repulsive electrostatic force that causes it to change direction. The magnitude of this force is given by Coulomb's law
F = k[tex]q_{1}[/tex][tex]q_{2[/tex] / [tex]r^{2}[/tex]
Where k is Coulomb's constant, [tex]q_{1}[/tex] and [tex]q_{2}[/tex] are the charges of the alpha particle and gold nucleus, respectively, and r is the distance between them. Since the alpha particle has a positive charge and the gold nucleus has a positive charge, the force is repulsive.
If we assume that the alpha particle is initially moving directly toward the center of the gold nucleus, then at the point of closest approach, the alpha particle will have a velocity v that is perpendicular to the direction from the alpha particle to the gold nucleus. At this point, the force on the alpha particle will be perpendicular to its velocity, which means that it will change only the direction of the alpha particle's velocity, not its magnitude.
Using conservation of momentum, we can relate the angle of deflection θ to the distance of closest approach r.
m[tex]v^{2}[/tex] / r = k[tex]q_{1}[/tex][tex]q_{2[/tex] / [tex]r^{2}[/tex]
Where m is the mass of the alpha particle. Solving for r, we get
r = k[tex]q_{1}[/tex][tex]q_{2[/tex] / m[tex]v^{2}[/tex]
To estimate the size of the gold nucleus, we assume that the alpha particles are deflected by a single, stationary gold nucleus at the center of the atom. In reality, the gold nucleus is not stationary, but this assumption gives us a rough estimate of its size.
Hence, the alpha particles are undeflected with a probability of 0.9993, we can assume that they do not interact with the gold nucleus and that their path is a straight line through the foil.
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Please help with the physics problem(single-loop circuits), provide step-by-step solution
Resistance of the resistors, R₁ and R₂ are 80Ω and 200Ω respectively.
V(A) = 12 V
ΔV(B) = 2 V
ΔV(C) = 5 V
R₃ = 200Ω
Since, the total voltage-drop along the upper branch must be 12 V, according to loop rule, the voltage-drop across resistor 3 is 5.0V
So, current through the loop,
I = V(C)/R₃
I = 5/200
I = 25 x 10⁻³A
a) Therefore, the resistance,
R₁ = V(B)/I
R₁ = 2/25 x 10⁻³
R₁ = 80Ω
b) Resistor 2 has the same voltage-drop as resistor 3. So, its resistance, R₂ is 200Ω.
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25.
The sensory hair cells of the semicircular canals convert vibrations into electrical impulses.
O True
False
It is a true statement that the sensory hair cells of the semicircular canals convert vibrations into electrical impulses.
How does sensory hair cells of the semicircular canals convert vibrations?The sensory hair cells located in the semicircular canals of the inner ear are responsible for converting mechanical vibrations caused by head movements into electrical impulses that are then transmitted to the brain for processing. This process helps the body maintain balance and spatial orientation.
The semicircular canals are part of the vestibular system in the inner ear, which plays a crucial role in detecting changes in head position and movement. The sensory hair cells in the semicircular canals are embedded in a gelatinous structure called the cupula, which moves in response to the flow of endolymphatic fluid inside the canals.
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Period, T [tex]\Longrightarrow T=\frac{60}{14} \Longrightarrow T \approx 4.286 \ s[/tex]
Using the equation, [tex]T=2 \pi \sqrt{\frac{l}{g} }[/tex], find l. ("g" is the acceleration from gravity, 9.8 m/s^2)
[tex]T=2 \pi \sqrt{\frac{l}{g} } \Longrightarrow 4.286=2 \pi \sqrt{\frac{l}{9.8} } \Longrightarrow \frac{4.286}{2 \pi} =\sqrt{\frac{l}{9.8} }[/tex]
[tex]\Longrightarrow (\frac{4.286}{2 \pi})^2 =\frac{l}{9.8} \Longrightarrow l =9.8(\frac{4.286}{2 \pi})^2 \Longrightarrow l \approx 4.560 \ m[/tex]
Thus, the radio booth is 4.560 m from the field.
Period, T [tex]\Longrightarrow T=\frac{60}{14} \Longrightarrow T \approx 4.286 \ s[/tex]
Using the equation, [tex]T=2 \pi \sqrt{\frac{l}{g} }[/tex], find l. ("g" is the acceleration from gravity, 9.8 m/s^2)
[tex]T=2 \pi \sqrt{\frac{l}{g} } \Longrightarrow 4.286=2 \pi \sqrt{\frac{l}{9.8} } \Longrightarrow \frac{4.286}{2 \pi} =\sqrt{\frac{l}{9.8} }[/tex]
[tex]\Longrightarrow (\frac{4.286}{2 \pi})^2 =\frac{l}{9.8} \Longrightarrow l =9.8(\frac{4.286}{2 \pi})^2 \Longrightarrow l \approx 4.560 \ m[/tex]
Thus, the radio booth is 4.560 m from the field.
Ten identical balls made of steel, each 27g are immersed in a measuring cylinder containing 20cm3 of water. The water level rises to a reading of 50cm3. What is the density of the steel?
Ten steel balls, each weighing 27g, are placed in a measuring cylinder filled with 20cm3 of water. The measurement of the water level increases to 50cm3.Then the density of the steel is 9 g/cm³.
Since the steel balls are fully immersed in water, the amount of water they displaced is equal to the volume of balls.
The volume of water before the balls were added is 20 cm³.
The water level rises to a reading of 50cm³ which means it is the total volume of water and the steel balls together. Therefore, the volume of the steel balls is:
50 cm³ - 20 cm³ = 30 cm³
It is given that there are ten identical balls, so the volume of each ball is:
30 cm³ ÷ 10 = 3 cm³
Each ball has a mass of 27 g.
So, the density of steel is calculated by following formula:
Density = Mass ÷ Volume
Density = 27 g ÷ 3 cm³
Density = 9 g/cm³
Therefore, the density of the steel balls is 9 g/cm³.
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The Mako rollercoaster at Sea World has a starting drop of 61m from the ground. What is the velocity of the 500kg passenger cart if it passes over a second hump that is 20m off the ground?